追加两个元素以使数组满足给定条件

给定非负整数的数组arr [] ，让我们将X定义为所有数组元素的XOR，将S定义为所有数组元素的总和。任务是找到两个元素，以便在将它们附加到数组后，对更新后的数组满足S = 2 * X。

例子：

Input: arr[] = {1, 7}
Output: 6 14
Initially S = 8, and X = 6. After appending 6
and 14, S_NEW = (8 + 6 + 14) = 28
and X_NEW = (6 ^ 6 ^ 14) = 14
Clearly, S_NEW = 2 * X_NEW

Input: arr[] = {1, 3}
Output: 2 6

天真的方法：运行两个从1到S的嵌套循环，并检查每对是否满足条件。这将花费O(S ² )时间。

高效方法：可以观察到，如果将X和S + X附加到数组，则满足给定条件的S_NEW = 2 *(S + X)和X_NEW = S +X 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to find the required numbers
void findNums(int arr[], int n)
{
  
    // Find the sum and xor
    int S = 0, X = 0;
    for (int i = 0; i < n; i++) {
        S += arr[i];
        X ^= arr[i];
    }
  
    // Print the required elements
    cout << X << " " << (X + S);
}
  
// Driver code
int main()
{
    int arr[] = { 1, 7 };
    int n = sizeof(arr) / sizeof(int);
  
    findNums(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
      
    // Function to find the required numbers 
    static void findNums(int arr[], int n) 
    { 
      
        // Find the sum and xor 
        int S = 0, X = 0; 
        for (int i = 0; i < n; i++) 
        { 
            S += arr[i]; 
            X ^= arr[i]; 
        } 
      
        // Print the required elements 
        System.out.println(X + " " + (X + S)); 
    } 
      
    // Driver code 
    public static void main (String[] args)
    { 
        int arr[] = { 1, 7 }; 
        int n = arr.length; 
      
        findNums(arr, n); 
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach 
  
# Function to find the required numbers 
def findNums(arr, n) : 
  
    # Find the sum and xor 
    S = 0; X = 0; 
    for i in range(n) : 
        S += arr[i]; 
        X ^= arr[i];
  
    # Print the required elements 
    print(X, X + S); 
  
# Driver code 
if __name__ == "__main__" : 
  
    arr = [ 1, 7 ]; 
    n = len(arr); 
  
    findNums(arr, n); 
      
# This code is contributed by AnkiRai01

C#

// C# implementation of the approach 
using System;
  
class GFG 
{
      
    // Function to find the required numbers 
    static void findNums(int []arr, int n) 
    { 
      
        // Find the sum and xor 
        int S = 0, X = 0; 
        for (int i = 0; i < n; i++) 
        { 
            S += arr[i]; 
            X ^= arr[i]; 
        } 
      
        // Print the required elements 
        Console.WriteLine(X + " " + (X + S)); 
    } 
      
    // Driver code 
    public static void Main()
    { 
        int []arr = { 1, 7 }; 
        int n = arr.Length; 
      
        findNums(arr, n); 
    } 
}
  
// This code is contributed by AnkitRai01

输出：

6 14

时间复杂度： O(n)